Limx → 1 (1-x) Tan (Wu X / 2)

Limx → 1 (1-x) Tan (Wu X / 2)

When x approaches to 1, find the limit of (1 + cos π x) / (x-1) ^ 2

Using the law of lobida
If LIM (f (x) / g (x)), X tends to a,
If the limits of F (x) and G (x) at a are both 0 or ∞, then
LIM (f (x) / g (x)) = LIM (f '(x) / g' (x)), X tends to a
This process can be continued, that is, if f '(x) and G' (x) also tend to 0 or ∞, then the second derivative can be continued
So when x tends to 1
lim(1+cosπx)/(x-1)^2
=lim(-πsinπx)/(2x-2)
=limit(-π^2cosπx/2)
=π^2/2

The limit of (e ^ x-1) / x ^ 2 when limx approaches 0

It's infinity
The Taylor expansion of (e ^ x-1) is (1 + X + 1 / 2x ^ 2 + 1 / 6x ^ 3 +...)
So you have 1 / X in your limit

Sum Sn = 2 times of 1 + 2x + 3x + 3 times of 3x +. + n-1 times of nx
Thank you!

Sn=1+2x+3x2+4x3+…… +nxn-1 ①
Multiply both sides by X at the same time
xSn=x+2x2+3x3+4x4+…… ＋nxn ②
① - 2
（1-x）Sn=1+X+X2+X3+…… +Xn-1-nXn
When x ≠ 1:
（1－x）Sn=(1-xn)/(1-x)-nxn
Sn=(1-xn)/(1-x)2-nxn/(1-x)
When x = 1
Sn=1+2+3+4+…… +n=n(n+1)/2
Note: X3 is the cubic power of X, nxn-1 is the power of N times X

Sum 1 + 2x + 3x ^ 2... + NX ^ n-1

Let Sn = 1 + 2x + 3x & # 178; +... + NX ^ (n-1)
Then xsn = x + 2x & # 178; + 3x & # 179; +... + (n-1) x ^ (n-1) + NX ^ n
Sn-xSn=(1-x)Sn=1+x+x²+...+x^(n-1)-nx^n=(1-x^n)/(1-x)-nx^n
Sn=(1-x^n)/(1-x)²-nx^n/(1-x)

Sum Sn = x + 2x2 + 3x3 + +nxn（x≠0）．

When x = 1, Sn = 1 + 2 + 3 + +When x ≠ 0 and X ≠ 1, Sn = x + 2x2 + 3x3 + +nxn，①xSn=x2+2x3+3x4+… +N xn + 1, and (1-x) Sn = x + x2 + X3 + +Xn-nxn + 1, so Sn = x (1 − xn) (1 − x) 2-nxn + 11 − X

How to use derivative summation method
For example, 1 ^ 2 + 2 ^ 2 + 3 ^ 2 How to sum n ^ 2

This sum does not need to use the derivative ah (n + 1); (n + 1) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\byadding all the above formulas, we can get: (n + 1

How to use derivative summation

Words of peace
For example, y = 1 / (x under 1-radical) + 1 / (x under 1 + radical)
=2/(1-X)
=2(1-X)^-1
Y'=(2(1-X)^-1)'
=2/(1-X)^2
In a word, four operation rules of formula derivative are used
Just pour it

Sn summation
The N-1 power of Sn = 1 + 2x3 + 3x9 + 4x27 +... + NX3

Note: m ^ n means the nth power of M, Sn = 1 + 2x3 + 3x3 ^ 2 + 4x3 ^ 3 + 5x3 ^ 4 +... + NX3 ^ (n-1) ① 3Sn = 3 + 2x3 ^ 2 + 3x3 ^ 3 + 4x3 ^ 4 +... + (n-1) X3 ^ (n-1) + NX3 ^ n, ② ① - ②, then - 2Sn = [1 + 3 + 3 ^ 2 + 3 ^ 3 + 3 ^ 4 +... + 3 ^ (n-1)] - NX3 ^ n = [3 ^ 0 +...]

An = n (n + 1), Sn =? Sequence summation
So strong? Let me do another one: A1 = 1 + 1 A2 = 1 / A + 4 A3 = 1 / A ^ 2 + 10 A (n) = 1 / A ^ (n-1) + (3n-2) for Sn

An = n (n + 1) = n ^ 2 + n Sn = (1 ^ 2 + 2 ^ 2 +... + n ^ 2) + (1 + 2 +... + n) = n (n + 1) (2n + 1) / 6 + n (n + 1) / 2 = n (n + 1) / 6 * [2n + 1 + 3] = n (n + 1) (n + 2) / 3 if you can use 1 ^ 2 + 2 ^ 2 +... + n ^ 2 = n (n + 1) (2n + 1) / 6, use the above. You can also do this. N (n + 1) = [n (n + 1) (n + 2) - n